well, I would say that $|A|=|A\cup B|$ means a bijection exists.

@LeakyNun Don't you agree with what Arthur and anonymous said ?

@Astyx I think I do.

The derivative of <c, f(x)> where f(x) is a vector valued function and c is a vector with real components is $<c, f^{'}(x)>$,correct?

Yes

Hi @Ted

Hi @Balarka

is it possible to get two different values for the determinant of a 3x3 matrix? The answers say 5 but I got -23

@q3d it isn't possible.

thanks

Otherwise it wouldn't be called the determinant :)

Hi @Ted

Hello, Zach.

How are you?

Doing ok, and are you better early afternoon than past midnight?

not exactly

Grr.

arxiv.org/pdf/1705.04152.pdf Look at this.

@PhysicsGuy: Most of us don't follow orders.

That's not an order. Look at it if you're interested.

Rehi @Ted and @Astyx

rehi Demonark

Rehi @Dami, long time no see (not)

What does Rehi mean?

Hi again

Sort of like repeat means to peat again.

Or restore to store

I understand.

Oh that never occurred to me

I like it

RE means recursively enumerable, in computability theory.

@Astyx Look at the Arxiv paper above. It's interesting.

I will after dinner if I find time - bye !

Demonark: It came from Latin, where there really was a re+petere :P

Can anyone explain the RHS of this image? physics-hell.tumblr.com/post/160052136531/…

Bon appétit, @Astyx

I think that for some reason I'm supposed to interpret "#R" as zero and I don't know why.

@EricStucky What?

what?

What?

What "what"?

Usually I interpret # to mean cardinality

Bingo

right, same

certainly infinite, @EricS: quadratics are zero compared to exponentials.

@EricStucky So, why should it be zero?

But I have no idea what they're doing re Euler's identity.

so 2^3j is an engineering joke:

(So many memes in one equation good lord)

e is about 2, pi is about 3, and engineers write j instead of i

Oh, physicists write $j$ for $i$, don't they?

so that's the e^pi*i bit

the summation is a math meme

meh ... garbage.

@TedShifrin Yes.

Blegh

because regularizing the zeta function at s=-1 gives \sum n = -1/12 (formally)

so the second term is the +1 bit

I do appreciate the name of this, the strongest Goldbach conjecture

I can only think of two common uses for $\#$. Either the cardinality of the set $A$ if $\#A$ is written, or the connected sum of manifolds $M$ and $N$ if $M\#N$ is written.

Have you all seen the various strength levels?

hi chat

so the RHS should be zero, I think

@Daminark any number is sum of two numbers, iirc?

hi @Eric

@Daminark isn't the strongest form of the goldbach conjecture that there are no numbers bigger than 7 or something

Yeah

lol

hmm steam, do you know how to construct the connect sum formally?

hi @Ted

In particular, is it true that $\varnothing \# X = \varnothing$ ?

oh i was thinking of very weak

that xkcd makes me laugh so hard

Two consecutive Erics, I am shocked.

You have to connect sum manifolds of the same dimension, in principle, @EricS.

you must be new here :P

hmm okay

Yes, I am quite new.

When there are two Ted's, this one will leave.

I think the connected sum is just undefined...

My friend and I have a meme going on about subscribing to the strongest Goldbach conjecture. $8$ is just infinity rotated so...

when there are two Balarka's, i'll kill the other

You have to remove open disks from each of the summands.

If there are two me's, I'll just give the other one half of my workload.

Sniped @Steamy

I've often joked before that it'd be nice to be able to divide oneself into various copies, have each one do very different things, and then all experiences reconverge

Hi DogAteMy

watch Dead Ringers

Ah, Jeremy Irons, @Balarka.

:)

You misapostrophed again, Balarka.

the tale is that Irons got an oscar for Dead Ringers, just on the next movie he did

Oops, following my having done so. hangs head in shame

haha, sorry Ted

That's why I don't talk about apostrophes, then noone can correct me.

Your syntax is usually correct, Mr. Bond.

In fact, I just talk rubbish, which isn't correctable.

This is the internet, anything can be corrected

Even correct things

especially correct things.

in Schlag's voice Precisely

For every correct thing there's a bunch of alternative facts :^)

Oh lawd

And also fake news.

nothing is correct, anything is permitted

How did they get the gradient in this problem?

@JingWeng Looks like it's from Wikipedia.

#helping

THis is the source; en.wikibooks.org/wiki/Undergraduate_Mathematics/…

Wow so inaccurate @Jason

But yeah this would follow from the chain rule

There cannot be two mes because no one had ever successfully impersonated my weirdness

I'm pretty sure there are at least two secrets on this website :P

@JingWeng: It's the multivariable chain rule.

$(f\circ\mathbf g)'(t) = \nabla f(\mathbf g(t))\cdot\mathbf g'(t)$

Does anyone else like Jacob Sartorius? I listen to his songs.

YEAAAAAAAAAAAAAAH

(... bad timing XD no I don't know who that is, Jason)

please

Well, he's the next Justin Bieber. You can find him on youtube.

not exactly

I listen to Jaco Pastorius, is that a good enough approximation?

Depending on who you talk to (including me) that's a huge indictment

@AlessandroCodenotti lol

"he's the next Justin Bieber" is a bug, not a feature

But most of all, Samy is my hero.

listen to death grips man. much better :P

re

@TedShifrin This is the multivariable chain rule that I know of: i.imgur.com/E1GYMvX.png. So $f(g(t))^{'}=f^{'}(g(t))g^{'}(t)=f^{'}(g(t))(y-x)$, how do I know this is equal to f(y)-f(x). I know by mean value therorem for 1d that $g(1)-g(0)=g^{'}(c)$ for some $c \in (0,1)$

Well not equal but leads to.

When $f$ is scalar-valued, you can write $f'(x)(v) = \nabla f(x)\cdot v$, @JingWeng.

@TedShifrin hi Ted

Hi, @Liad. I'm in the middle of trying to solve a problem on main.

Alright, if you'll have time latter ping me :P

OK, @Liad. What's up?

@TedShifrin Now great :P

Zut! @Hippa!

@TedShifrin o/

@TedShifrin we learned about conservative vector field , and i got a question

Yeah?

im not sure about the term in English, is there a subject called "local conservative" ?

the TA defined it when the crossed partial derivatives equals

@Liad: Yeah, that's a theorem (called the Poincaré lemma). If you have mixed partials equal, then it's locally exact.

alright

so if we have the mixed partials equal then the function is locally exact

And you should know an example where it's locally conservative but not conservative.

well, the differential is locally exact, yeah.

yea i know an example of that

OK, so what's the question?

the definition we had for conservative vector field was that each closed cuve has integral 0

does this implies locally exact?

Right, ok. No, that implies globally exact.

one is global one is local so i guess yes, but i dont see it from the def.

So exact means that $\vec F = \nabla f$ for some potential function $f$.

globally exact does not imply local?

Of course, global implies local.

huh, you wrote "No," so i thought im mistaken

There's a theorem here. Integral 0 around all closed curves is equivalent to $\vec F =\nabla f$.

No meaning, not /just/ locally exact

I said no to emphasize that there's something more powerful going on. But it does imply what you said, yes.

Huh, so if a vector field is globally exact , it must be a gradient of some F ?

we learned if it is a gradient , then it is globally exact

That's tautology to me. Saying "exact" means it's a gradient.

I thought I'd seen all linear algebra problems. This one was new to me. @Astyx @Daminark

alright

now for another thing , if $T: \Bbb R \ ^ n \to \Bbb R \ ^ n$ defined by $T(x) = Ax +b$ where $A$ is orthogonal matrix

i think the derivative of $T $ is $A$

Sure.

Oh huh, this is interesting

yea it is easy , but when i wanted to show it formally , i kind of forgot how to do it by definition :P

It relates to affine geometry, Demonark. Where you look at the affine subspace spanned by vectors $v_1,\dots,v_k$. That turns out to be all linear combinations $\sum c_iv_i$ where $\sum c_i=1$.

Well, write down the limit that the derivative must satisfy and show it works, @Liad. It's easy.

i need to show that $\dfrac{||T(x+h) - T(x) ||}{||h||}$ tends to zero when $||h||$ does?

because that expression is always one :P

no, @Liad, you need $T(x+h)-T(x)-DT(x)(h)$ in the numerator.

And you're guessing that $DT(x) = A$.

Huhh...

i knew it

yea , great, it is always zero now :P

Yup.

thanks

Sure.

Nice

i want to make sure , if i show that the mixed derivatives is not equals - that means the function is not locally/globally exact, correct?

Woops. I read too fast.

:P

The correct statement is mixed partials equal if and only if locally exact.

So, yes, I was right and you were right.

But mixed partials equal does NOT imply conservative (globally exact).

alright, you should replace my TA , he made a mess in class ..

All this stuff is in my lectures on YouTube, but I did it all using differential forms, which you probably are not doing.

Woo forms

A lot of mathematicians (and grad students in particular) do not use this kind of mathematics and do not know it well.

I've started one section of GP which we skipped in class, about De Rham cohomology, and the stuff is really cool

So I'm starting to like the forms camp

Yeah, awesomely cool. I always proved Mayer-Vietoris for deRham in class and computed cohomology of spheres, then did top cohomology of manifolds.

I can send you those pages of my notes, Demonark, but they don't have full details.

if we end up making a crew then we could all do forms

You're falling behind, @Balarka. :P

Actually one of our analysis problems last quarter, I've now realized, was computing that of the punctured plane

Right, Demonark. I think we talked about that.

nah i think i know the basics

things like punctured plane etc can just be done via vector analysis, so it isn't much inspiring to me to do it with forms tbh

Cool @Ted

@TedShifrin i always wanted to ask a professor :P . was it easy for you the first degree in math. ?

@Balarka Vector analysis meaning, classical stuff?

Curl, Divergence, etc

Yeah, @Liad, pretty much. Although Michael Artin gave me an A in Algebra I most definitely did not deserve. The other interesting story (which I've told my students many times) is that I got my one B in the course on multivariable analysis (and Lebesgue integrals) — since I've now written a book doing the material, you might say I know it reasonably well. :P

Right. Integrating 1-forms along path is just line integral of the vector field it's dual to.

And yes, Stokes <=> Green

@Balarka: If I send Demonark my notes, you can help him with Mayer-Vietoris for DeRham. It uses partitions of unity. Very cool.

Irrotational vector fields are closed forms, conservative vector fields are exact

You can write down the maps so explicitly for DeRham.

Oh @Ted I think I'd very much appreciate that! :D

@TedShifrin Hm, OK.

I can try.

I worked my butt off, @Liad, but I took a lot of graduate courses, too, when I was an undergraduate.

@Balarka: I'll send you those pages, too.

Thanks.

Demonark, do I have your email?

Demonark, did you guys prove Poincaré-Hopf?

I believe so, though I can send you one again if you'd prefer

Yeah, I found it, Demonark.

Not yet, though Neves intends to do so soon

He said the last few weeks of class will be applications

OK, so I won't send that. I'll just send the forms stuff.

Hey @Mike!

My notes get scribbly because at the end of my course is when the cancer got bad and I was totally weak (and actually missed the last two weeks of classes).

hi

How's it going?

Morning

g'night, @MikeM.

And ouch @Ted, that sucks

I need to know what a azeotropic mixture is before I start mathing tho

Wait night?

it's always night for mike

OK, sent.

Awesome, thanks!

And lel

received

Oh, so there's one problem in GP which says that defines a closed curve and line integral on a manifold. Now, it says that if your curve is in $\mathbb{R}^k$, to write the integral explicitly in terms of coordinate representations of $\gamma$ and $\omega$

Is the coordinate representation just $\omega = \sum_{i=1}^k a_idx_i$ and $\gamma = (\gamma_1,\ldots,\gamma_k)$?

Or is it asking something else?

Yup.

ya

Pull back explicitly.

(Or look in my book ... Ha.)

1-forms are super ez

True, I am happy that we're seeing the full formulation finally

snaps

That's as bad as my footnote about the form that forms take.

Oh I remember that :P

Actually, not even a footnote. The footnote was just a faux-apology to rub it in.

I smiled a little when I read it.

That was before you became a ... um ... sophisticated savant re puns, @Balarka.

Demonark, did you get the email?

Yeah, I did, thanks!

OK, just checking.

There's more stuff on forms (e.g., the Poincaré lemma proof) in there, too.

But Mayer-Vietoris is super-cool.

@TedShifrin "When they were jung and easily freudened" - James Joyce.

Computing cohomology of a manifold by computing the cohomology of two open pieces whose union is the whole thing (and using the intersection, as well).

Oy @Balarka.

rofl

Explicit function theorem: Let $f :\Bbb R^n \to \Bbb R^m$ be a continuously differentiable function and let $U\subset \Bbb R^n$ be an open set. Let $y = f(x)$ for $x\in U$. Then there exists a continuously differentiable function $\phi : \Bbb R^{n+m} \to \Bbb R^m$ and a constant vector $c\in\Bbb R^m$ such that $\phi(x,y) = c$ for $x\in U$.

Huh?

O lawd

explicit? lol

It's my contribution to maths

Nice, lel

Hi @TedShifrin

and Balarka!

Hi Danu

Hey @Danu!

I found out something funny today. I can only compute the cohomology ring & Chern classes of a certain manifold up to a single sign which I can't seem to pin down. But it doesn't matter for the Chern numbers, because things conspire so that the signs all cancel anyways :D

Hi Danu

hi Astyx and Daminark

Hi/bye @Danu. Leaving for lunch.

@TedShifrin Oh, okay.l

See you @Ted!

Back later.

See you later

Bye @Ted, bon appétit

@blue Are you here?

Oh and hi @Waiting

@blue If I have a solution which deviates positively or negatively from an ideal solution, I get a maximum resp. minimum vapor pressure at a particular composition. So for this composition, the mixture has lowest resp. highest boiling point. Why's it an azeotropic mixture at this composition?

Lol I looked at Artin and one thing in the appendix is the implicit function theorem

This is... interesting...

@Daminark Cool, eh?

@blue Thanks, let me look.

I wonder where it uses it in the book

I see that they say that it has to be first order, but still.. it has to make sense, no?

@Daminark Possibly where he explains Riemann surfaces.

... he does Riemann surfaces?

@Sha That's not linear

Yep, as a geometric motivation for field theory and extensions.

@Astyx I see, but we can't just drop the squares and square root, right? Or is it like the best linear approximation we can make?

Field extensions of C(z) correspond to Riemann surfaces (branched over P^1)

Having a square root and squares doesn't even make sense?

What if $f$ decreases?

I think you're confusing some things. Here $df(a)$ is a vector, not a distance

$df$ is a scalar

you can express it as a dot product

and @Steamy that is indeed a good point

Oh right then you're acting as if the changes in $x,y,z$ were orthogonal

so $df\neq \vert d\vec f\vert$...?

It's just a variation along a line

Nice

Like, if you change $x$ a little you go from 0 to ${\partial f\over \partial x} dx$

yea

but it's 3D, no?

oh, wait

Not f, as you said, it's a scalar

ahh yes i see

I think

right right

Maybe think of it as the altitude on a hill

thanks

Oh I've been thinking at some point I want to teach a class such that in the first half of the quarter I talk in a different accent

yea, exactly

lol :P

If you move a little bit along x, then along y, the variation of the altitude will be the sum of the variations

yeaa I get it @Astyx

i confused it as a vector indeed

Halfway through I revert and everyone's just like wat?

I like that.

@Astyx bonsoir

Bonsoir

je me suis arrêté ce matin au fait que $d(x,y)\in ]d(x,y)-r,d(x,y)+r[\subset U$ et donc $(x,y)\in d^{-1}(]d(x,y)-r, d(x,y)+r[\subset d^{-1}(U)$

@Astyx

2 minutes

ok merci

Pose $B$ la boule de rayon $r'$ centrée sur $(x,y)$ @Vrouvrou

What would you guys say are books whose titles made you think the book was about something that it very much isn't about? "Basic Number Theory" came to me as quite the surprise, fit example

Rehi Ted

re @Astyx

Rehi @Ted!

re Demonark

Well, this is sure thrilling ...

Probability textbook uses E_i to denote event that ith experiment succeeds. It then says that if for all i_1, i_2, ..., i_n holds equality P(E_i_1, E_i_2, ..., E_i_n) = P(E_i_1)P(E_i_2)...P(E_i_n), we say that the sequence of experiments consists of independent trials.

I want to understand why double indexing. Why cannot I say if for i in {1, 2, ..., n} equality P(E_1, E_2, ..., E_n) = P(E_1)P(E_2)...P(E_n) holds then we have independent trials?

As if on queue

@Astyx centré en $d(x,y)$ ?

Non, en $(x,y)$

@valentin: You need independence of pairs, triples, quadruples, etc.

Si tu arrives à montrer que don image est incluse dans celle de rayon $r$ centrée en $d(x,y)$, c'est gagné

How is that related?

You're thinking that if you have the one $n$-fold independence equation, that all the others will follow. Surely that isn't true, but I don't have an example off-hand.

Don't they have a different subscript on $i_1,\dots,i_k$, rather than $i_1,\dots,i_n$? I agree that if it's $n$, there's no difference.

Or maybe $n$ is varying from $2$ to ... the number of trials.

Or maybe in fact there are infinitely many trials ... so you have to look all possible finite subcollections.

@Astyx Hi

mutters something about finite measure spaces

@Astyx je suis désolé mais je n'ai toujours pas compris

s'il vous plait

@valentin: I think that they're allowing infinitely many trials. So you have to talk about independence of every finite subset.

@Astyx s'il vous plait je ne sais plus ce que je doit montrer

@TedShifrin Here is original text goo.gl/B3Nu1u

How do I get subsets there?